#### Answer

$$\frac{a}{3}$$

#### Work Step by Step

We have the derivative:
$$
f^{\prime}(x)=3x^{2}
$$
Then at $x=a, m=f^{\prime}(a)=3a^{2}$. Hence, the tangent line is:
$$
\begin{aligned}
\frac{y-y_{1}}{x-x_{1}} &=m \\
\frac{y-a^n}{x-a} &=3a^{2} \\
y &=3a^{2} (x-a)+a^3
\end{aligned}
$$
Since the tangent line intersects with the $x-$ axis at $x=0,$ then $Q$ has coordinates $(a-\frac{a}{3},0), R$ has coordinates $(a,0)$, and the subtangent is
$$
a-\left(a-\frac{a}{3}\right)=\frac{a}{3}
$$